Visibility of the horizon line. Navigation theory
Visible horizon. Considering that the earth's surface is close to a circle, the observer sees this circle bounded by the horizon. This circle is called the visible horizon. The distance from the location of the observer to the visible horizon is called the range of the visible horizon.
It is extremely clear that the higher above the ground (surface of water) the observer's eye is located, the greater will be the range of the visible horizon. The range of the visible horizon at sea is measured in miles and is determined by the formula:
where: De - range of the visible horizon, m;
e is the height of the observer's eye, m (meter).
To get the result in kilometers:
Range of visibility of objects and lights. Visibility range object (a lighthouse, another ship, a structure, a rock, etc.) at sea depends not only on the height of the observer's eye, but also on the height of the observed object ( rice. 163).
Rice. 163. Beacon visibility range.
Therefore, the visibility range of the object (Dn) will be the sum of De and Dh.
where: Dn - visibility range of the object, m;
De - range of the visible horizon by the observer;
Dh - range of the visible horizon from the height of the object.
The visibility range of an object above the water level is determined by the formulas:
Dp = 2.08 (√е + √h), miles;
Dp = 3.85 (√е + √h), km.
Example.
Given: the height of the navigator's eye e = 4 m, the height of the lighthouse h = 25 m. Determine at what distance the navigator should see the lighthouse in clear weather. Dp = ?
Solution: Dp = 2.08 (√e + √h)
Dp = 2.08 (√4 + √25) = 2.08 (2 + 5) = 14.56 m = 14.6 m.
Answer: The lighthouse will open to the observer at a distance of about 14.6 miles.
On practice skippers the visibility range of objects is determined either by a nomogram ( rice. 164), or according to nautical tables, using maps, sailing directions, descriptions of lights and signs. You should be aware that in the manuals mentioned, the visibility range of objects Dk (card visibility range) is indicated at the height of the observer's eye e = 5 m and, in order to obtain the true range of a particular object, it is necessary to take into account the correction DD for the difference in visibility between the actual height of the observer's eye and the card height e = 5 m. This problem is solved with the help of nautical tables (MT). The determination of the visibility range of an object according to the nomogram is carried out as follows: the ruler is applied to known values the height of the eye of the observer e and the height of the object h; the intersection of the ruler with the average scale of the nomogram gives the value of the desired value Dn. On fig. 164 Dp = 15 m with e = 4.5 m and h = 25.5 m.
Rice. 164. Nomogram for determining the visibility of an object.
When studying the issue of visibility range of lights at night it should be remembered that the range will depend not only on the height of the fire above the sea surface, but also on the strength of the light source and on the type of lighting apparatus. As a rule, the lighting apparatus and lighting strength are calculated for lighthouses and other navigational signs in such a way that the visibility range of their lights corresponds to the visibility range of the horizon from the height of the light above sea level. The navigator must remember that the visibility range of an object depends on the state of the atmosphere, as well as topographic (color of the surrounding landscape), photometric (color and brightness of the object against the background of the terrain) and geometric (size and shape of the object) factors.
An observer, being at sea, can see one or another landmark only if his eye is above the trajectory or, in the limiting case, on the very trajectory of the beam coming from the top of the landmark tangent to the surface of the Earth (see figure). Obviously, the mentioned limiting case will correspond to the moment when the landmark is revealed to the observer approaching it or hidden when the observer moves away from the landmark. The distance on the surface of the Earth between the observer (point C), whose eye is at point C1 and the object of observation B with the apex at point B1 corresponding to the moment of opening or hiding this object, is called the visibility range of the landmark.
The figure shows that the range of visibility of the landmark B is the sum of the range of the visible horizon BA from the height of the landmark h and the range of the visible horizon AC from the height of the observer's eye e, i.e.
Dp \u003d arc BC \u003d arc BA + arc AC
Dp \u003d 2.08v h + 2.08v e \u003d 2.08 (v h + v e) (18)
The visibility range calculated by formula (18) is called the geographical visibility range of an object. It can be calculated by adding selected from the above table. 22-a MT separately for the range of the visible horizon for each of the given heights h u e
According to the table 22nd we find Dh=25 miles, De=8.3 miles.
Consequently,
Dp \u003d 25.0 +8.3 \u003d 33.3 miles.
Tab. 22-c, placed in the MT, makes it possible to directly obtain the full range of visibility of the landmark from its height and the height of the observer's eye. Tab. 22-c is calculated by formula (18).
You can see this table here.
On the nautical charts and navigation aids show the visibility range Dn of landmarks for a constant height of the observer's eye equal to 5 m. The range of opening and hiding objects in the sea for an observer whose eye height is not equal to 5 m will not correspond to the visibility range Dk shown on the map . In such cases, the visual range of landmarks shown on the map or in manuals must be corrected by a correction for the difference between the height of the observer's eye and the height, equal to 5 m. This correction can be calculated based on the following considerations:
Dp \u003d Dh + De,
Dk \u003d Dh + D5,
Dh \u003d Dk - D5,
where D5 is the range of the visible horizon for the observer's eye height of 5 m.
Substitute the value of Dh from the last equality into the first:
Dp \u003d Dk - D5 + De
Dp = Dk + (De - D5) = Dk + ^ Dk (19)
The difference (De - D5) \u003d ^ Dk and is the desired correction to the visibility range of the landmark (fire) indicated on the map, for the difference between the height of the observer's eye and the height equal to 5 m.
For convenience on a cruise, it is possible to recommend to the navigator to have on the bridge corrections calculated in advance for different levels of the eye of an observer located on various superstructures of the ship (deck, navigation bridge, signal bridge, gyrocompass pelorus installation sites, etc.).
Example 2. The map near the lighthouse shows the visibility range Dk = 18 miles. Calculate the visibility range Dp of this lighthouse from an eye height of 12 m and the height of the lighthouse h.
According to the table 22nd MT we find D5 = 4.7 miles, De = 7.2 miles.
We calculate ^ Dk \u003d 7.2 - 4.7 \u003d + 2.5 miles. Consequently, the visibility range of a beacon with e = 12 m will be equal to Dp = 18 + 2.5 = = 20.5 miles.
By the formula Dk = Dh + D5 we define
Dh \u003d 18 - 4.7 \u003d 13.3 miles.
According to the table 22nd MT by the return entrance we find h = 41 m.
Everything stated about the range of visibility of objects in the sea refers to the daytime, when the transparency of the atmosphere corresponds to its average state. During transitions, the navigator must take into account possible deviations of the state of the atmosphere from the average conditions, accumulate experience in assessing visibility conditions in order to learn to anticipate possible changes in the visibility range of objects at sea.
At night, the visibility range of beacon lights is determined by the optical visibility range. The optical range of visibility of the fire depends on the strength of the light source, on the properties of the optical system of the beacon, the transparency of the atmosphere and on the height of the installation of the fire. The optical range of visibility may be greater or less than the daytime visibility of the same beacon or light; this range is determined experimentally from multiple observations. The optical range of visibility of beacons and lights is selected for clear weather. Typically, light-optical systems are selected so that the optical and daytime geographical visibility ranges are the same. If these ranges differ from one another, then the smaller of them is indicated on the map.
The range of visibility of the horizon and the range of visibility of objects for the real atmosphere can be determined empirically using a radar station or by observations.
Each object has a certain height H (Fig. 11), therefore, the visibility range of the object Dp-MR is composed of the range of the visible horizon of the observer De=Mc and the range of the visible horizon of the object Dn=RC:
Rice. eleven.
According to formulas (9) and (10), H. N. Struisky compiled a nomogram (Fig. 12), and in MT-63, Table. 22-c "Visibility range of objects", calculated by formula (9).
Example 11. Find the range of visibility of an object with a height above sea level H = 26.5 m (86 feet) at the height of the observer's eye above sea level e = 4.5 m (15 feet).
Solution.
1. According to the Struisky nomogram (Fig. 12), on the left vertical scale "Height of the observed object" we mark the point corresponding to 26.5 m (86 ft), on the right vertical scale "Height of the observer's eye" we mark the point corresponding to 4.5 m ( 15 ft); connecting the marked points with a straight line, at the intersection of the latter with the average vertical scale "Visibility range" we get the answer: Дn = 15.1 m.
2. According to MT-63 (Table 22-c). For e = 4.5 m and H = 26.5 m, the value Dn = 15.1 m. the observer's eye is not equal to 5 m, then it is necessary to add the correction A \u003d MS-KS- \u003d De-D5 to the range Dk given in the manuals. The correction is the difference between the distances of the visible horizon from a height of 5 m and is called the correction for the height of the observer's eye:
As can be seen from formula (11), the correction for the height of the observer's eye A can be positive (when e > 5 m) or negative (when e
So, the visibility range of a beacon light is determined by the formula
Rice. 12.
Example 12. Beacon visibility range indicated on the map, Dk = 20.0 miles.
From what distance can an observer see the fire, whose eye is at a height of e = 16 m.
Solution. 1) by formula (11)
2) according to the table. 22-a ME-63 A \u003d De - D5 \u003d 8.3-4.7 \u003d 3.6 miles;
3) according to the formula (12) Dp \u003d (20.0 + 3.6) \u003d 23.6 miles.
Example 13 The visibility range of the beacon, indicated on the map, Dk = 26 miles.
From what distance will the observer on the boat see the fire (e = 2.0 m)
Solution. 1) by formula (11)
2) according to the table. 22-a MT-63 A = D - D = 2.9 - 4.7 = -1.6 miles;
3) according to the formula (12) Dp = 26.0-1.6 = 24.4 miles.
The visibility range of an object, calculated by formulas (9) and (10), is called geographical.
Rice. 13.
Visibility range of a beacon light, or optical range visibility depends on the strength of the light source, the beacon system and the color of the fire. In a well-built lighthouse, it usually coincides with its geographical range.
In cloudy weather, the actual visual range may differ significantly from the geographic or optical range.
AT recent times research has established that in daytime sailing conditions, the visibility range of objects is more accurately determined by the following formula:
On fig. Figure 13 shows the nomogram calculated by formula (13). We will explain the use of the nomogram by solving a problem with the conditions of example 11.
Example 14 Find the range of visibility of an object with a height above sea level H = 26.5 m, with the height of the observer's eye above sea level e = 4.5 m.
Solution. 1 by formula (13)
Geographic range visibility of objects in the sea D p is determined by the greatest distance at which the observer will see its top above the horizon, i.e. depends only on the geometric factors that relate the height of the observer's eye e and the height of the landmark h at a refractive index c (Fig. 1.42):
where D e and D h - respectively, the range of the visible horizon from the height of the observer's eye and the height of the object. That. the range of visibility of an object, calculated from the height of the observer's eye and the height of the object is called geographic or geometric range of visibility.
The calculation of the geographical range of visibility of an object can be made according to Table. 2.3 MT - 2000 according to the arguments e and h or according to the table. 2.1 MT - 2000 by summing the results obtained by double entry into the table for the arguments e and h. You can also get D p according to the Struisky nomogram, which is given in MT - 2000 under the number 2.4, as well as in each book "Lights" and "Lights and Signs" (Fig. 1.43).
On nautical navigation charts and in navigation manuals, the geographical range of visibility of landmarks is given for a constant height of the observer's eye e = 5 m and is denoted as Dk - the visibility range indicated on the map.
Substituting the value e = 5 m into the formula (1.126), we get:
To determine D p, it is necessary to introduce an amendment D to D to, the value of which and the sign are determined by the formula:
If the actual height of the eye is more than 5 m, then DD has a “+” sign, if it is less, a “-” sign. In this way:
. (1.129)
The value of D p also depends on visual acuity, which is expressed in the resolution of the eye in terms of angle, i.e. it is also determined by the smallest angle at which the object and the horizon line differ separately (Fig. 1.44).
According to formula (1.126)
But due to the resolving power of the eye g, the observer will see the object only when its angular dimensions are not less than g, i.e. when it is visible above the horizon line at least Dh, which from the elementary DA¢CC¢ at angles C and C¢ close to 90° will be Dh = D p × g¢.
To get D p g in miles with Dh in meters:
where D p g - geographical range of visibility of the object, taking into account the resolution of the eye.
Practical observations have determined that when the beacon is opened, g =2¢, and when hidden, g =1.5¢.
Example. Find the geographical range of visibility of a beacon with a height of h=39 m, if the height of the observer's eye is e=9 m, without taking into account and taking into account the resolution of the eye g = 1.5¢.
Influence of hydrometeorological factors on the visibility range of lights
In addition to geometric factors (e and h), the visibility range of landmarks is also affected by contrast, which makes it possible to distinguish a landmark from the surrounding background.
The range of visibility of landmarks during the day, which also takes into account the contrast, is called daytime optical range of visibility.
To ensure safe navigation at night, special means of navigational equipment are used with light-optical devices: beacons, luminous navigation signs and navigation lights.
Marine Lighthouse - this is a special permanent structure with a visibility range of white or colored lights brought to it at least 10 miles.
Glowing maritime navigation mark- a capital structure with a light-optical apparatus with a visibility range of white or colored lights brought to it less than 10 miles.
Marine navigational light- a light device installed on natural objects or structures of non-special construction. Such aids to navigation often operate automatically.
At night, the visibility range of beacon lights and luminous navigation signs depends not only on the height of the observer's eye and the height of the luminous AtoN, but also on the strength of the light source, the color of the fire, the design of the light-optical apparatus, and also on the transparency of the atmosphere.
The visibility range that takes into account all these factors is called night optical range of visibility, those. is the maximum visual range of the fire at a given time for a given meteorological visual range.
Meteorological visibility range depends on the transparency of the atmosphere. Part of the luminous flux of lights of luminous aids to navigation is absorbed by particles contained in the air, therefore, there is a weakening of the luminous intensity, characterized by atmospheric transparency coefficient t:
where I 0 - luminous intensity of the source; I 1 - light intensity at a certain distance from the source, taken as a unit (1 km, 1 mile).
The transparency coefficient of the atmosphere is always less than one, so the geographical visibility range is usually greater than the real one, except in anomalous cases.
The transparency of the atmosphere in points is estimated according to the visibility scale of the table 5.20 MT - 2000, depending on the state of the atmosphere: rain, fog, snow, haze, etc.
Since the optical range of lights varies considerably with the transparency of the atmosphere, the International Association of Lighthouse Authorities (IALA) has recommended the use of the term “nominal visual range”.
Nominal visual range of fire is called the optical range of visibility at a meteorological visibility range of 10 miles, which corresponds to the transparency coefficient of the atmosphere t = 0.74. The nominal range of visibility is indicated in the navigation manuals of many foreign countries. On the domestic maps and the sailing manuals state the standard visual range (if less than the geographic visual range).
Standard line of sight fire is called the optical range of visibility at a meteorological visibility range of 13.5 miles, which corresponds to the transparency coefficient of the atmosphere t = 0.8.
In the navigation aids “Lights”, “Lights and signs”, in addition to the table of the range of the visible horizon and the nomogram of the visibility range of objects, there is also a nomogram of the optical visibility range of lights (Fig. 1.45). The same nomogram is given in MT - 2000 under the number 2.5.
The arguments for entering the nomogram are the luminous intensity, or nominal or standard visual range, (obtained from navigation aids), and the meteorological visual range, (obtained from the meteorological forecast). According to these arguments, the optical range of visibility is obtained from the nomogram.
When designing beacons and lights, they strive for the optical range of visibility to be equal to the geographical range of visibility in clear weather. However, for many lights the optical range is less than the geographic range. If these ranges are not equal, charts and sailing manuals indicate the smaller of them.
For practical calculations of the expected visual range of fire afternoon it is necessary to calculate D p according to the formula (1.126) from the heights of the observer's eye and the landmark. At night: a) if the optical visibility range is greater than the geographical one, it is necessary to take a correction for the height of the observer's eye and calculate the geographical visibility range using formulas (1.128) and (1.129). Take the smaller of the optical and geographical, calculated by these formulas; b) if the optical range of visibility is less than the geographical one, take the optical range.
If on the map near the fire or lighthouse D to< 2,1 h + 4,7 , то поправку DД вводить не нужно, т.к. эта дальность видимости оптическая меньшая географической дальности видимости.
Example. The height of the observer's eye e = 11 m, the visibility range of the fire indicated on the map D k = 16 miles. The nominal range of visibility of the beacon from the navigation manual "Lights" is 14 miles. Meteorological visibility range 17 miles. At what distance can we expect the lighthouse to open fire?
According to the nomogram Dopt » 19.5 miles.
By e \u003d 11m ® D e \u003d 6.9 miles
D 5 = 4.7 miles
DD =+2.2 miles
D to = 16.0 miles
D p \u003d 18.2 miles
Answer: Fire can be expected from a distance of 18.2 miles.
Nautical charts. Map projections. Gaussian transverse conformal cylindrical projection and its use in navigation. Perspective projections: stereographic, gnomonic.
A map is a reduced distorted image of the spherical surface of the Earth on a plane, provided that the distortions are regular.
Plan - an image not distorted due to the smallness of the depicted area earth's surface on surface.
Cartographic grid - a set of lines depicting meridians and parallels on the map.
Map projection is a mathematically based way of depicting meridians and parallels.
A geographic map is a conditional image of the entire earth's surface or part of it built in a given projection.
Maps are different in purpose and scale, for example: planispheres - depicting the entire Earth or a hemisphere, general or general - depicting individual countries, oceans and seas, private - depicting smaller spaces, topographic - depicting details of the land surface, orographic - relief maps, geological - the occurrence of layers, etc.
Nautical charts are special geographical charts designed primarily to provide navigation. AT general classification geographical maps they are classified as technical. Special place among sea charts occupy MNCs, which serve to lay the course of the vessel and determine its place in the sea. The ship's collection may also contain auxiliary and reference charts.
Classification of cartographic projections.
According to the nature of the distortions, all cartographic projections are divided into:
- Equangular or conformal - projections in which the figures on the maps are similar to the corresponding figures on the surface of the Earth, but their areas are not proportional. The angles between objects on the ground correspond to those on the map.
- Equal-sized or equivalent - in which the proportionality of the areas of the figures is preserved, but the angles between the objects are distorted.
- Equidistant - preserving the length along one of the main directions of the distortion ellipse, i.e., for example, a circle on the ground on the map is depicted as an ellipse, in which one of the semi-axes is equal to the radius of such a circle.
- Arbitrary - all the rest that do not have the above properties, but are subject to other conditions.
According to the method of constructing the projection, they are divided into:
|
Depending on the point of contact of the picture plane, gnomonic projections are divided into: normal or polar - touching on one of the poles transverse or equatorial - touching - on the equator
horizontal or oblique - touching at any point between the pole and the equator (the meridians on the map in such a projection are rays diverging from the pole, and the parallels are ellipses, hyperbolas or parabolas.
Horizon visibility range
The line observed in the sea, along which the sea, as it were, connects with the sky, is called visible horizon of the observer.
If the observer's eye is at a height eat above sea level (ie. BUT rice. 2.13), then the line of sight going tangentially to the earth's surface defines a small circle on the earth's surface aa, radius D.
Rice. 2.13. Horizon visibility range
This would be true if the Earth were not surrounded by an atmosphere.
If we take the Earth as a sphere and exclude the influence of the atmosphere, then from right triangle OAa follows: OA=R+e
Since the value is extremely small ( for e = 50m at R = 6371km – 0,000004 ), then we finally have:
Under the influence of the earth's refraction, as a result of the refraction of the visual beam in the atmosphere, the observer sees the horizon further (in a circle centuries).
(2.7)
where X- coefficient of terrestrial refraction (» 0.16).
If we take the range of the visible horizon D e in miles, and the height of the observer's eye above sea level ( eat) in meters and substitute the value of the Earth's radius ( R=3437,7 miles = 6371 km), then we finally obtain a formula for calculating the range of the visible horizon
(2.8)
For example: 1) e = 4 m D e = 4,16 miles; 2) e = 9 m D e = 6,24 miles;
3) e = 16 m D e = 8,32 miles; 4) e = 25 m D e = 10,4 miles.
According to formula (2.8), table No. 22 "MT-75" (p. 248) and table No. 2.1 "MT-2000" (p. 255) according to ( eat) from 0.25 m¸5100 m. (see table 2.2)
Range of visibility of landmarks at sea
If an observer whose eye height is at a height eat above sea level (ie. BUT rice. 2.14), observes the horizon line (i.e. AT) on distance D e(miles), then, by analogy, and from a landmark (i.e., B), whose height above sea level hM, visible horizon (ie. AT) is observed at a distance Dh(miles).
Rice. 2.14. Range of visibility of landmarks at sea
From fig. 2.14 it is obvious that the range of visibility of an object (landmark) having a height above sea level hM, from the height of the observer's eye above sea level eat will be expressed by the formula:
Formula (2.9) is solved using table 22 "MT-75" p. 248 or Table 2.3 "MT-2000" (p. 256).
For example: e= 4 m, h= 30 m, D P = ?
Solution: for e= 4 m® D e= 4.2 miles;
for h= 30 m® D h= 11.4 miles.
D P= D e + D h= 4,2 + 11,4 = 15.6 miles.
Rice. 2.15. Nomogram 2.4. "MT-2000"
Formula (2.9) can also be solved using Apps 6 to "MT-75" or nomograms 2.4 "MT-2000" (p. 257) ® fig. 2.15.
For example: e= 8 m, h= 30 m, D P = ?
Solution: Values e= 8 m (right scale) and h\u003d 30 m (left scale) we connect with a straight line. The point of intersection of this line with the average scale ( D P) and gives us the desired value 17.3 miles. ( see table. 2.3 ).
Geographic range of visibility of objects (from Table 2.3. "MT-2000")
Note:
The height of the navigational landmark above sea level is selected from the navigational manual for navigation "Lights and Signs" ("Lights").
2.6.3. Range of visibility of the landmark light shown on the map (Fig. 2.16)
Rice. 2.16. Beacon light visibility ranges shown
On nautical nautical charts and in navigation aids, the range of visibility of the landmark light is given for the height of the observer's eye above sea level. e= 5 m, i.e.:
If the actual height of the observer's eye above sea level differs from 5 m, then to determine the visibility range of the landmark fire, it is necessary to add to the range shown on the map (in the manual) (if e> 5 m), or subtract (if e < 5 м) поправку к дальности видимости огня ориентира (DD K) shown on the map for the height of the eye.
(2.11)
(2.12)
For example: D K= 20 miles, e= 9 m.
D O = 20,0+1,54=21,54miles
then: DO = D K + ∆ D To = 20.0+1.54 =21.54 miles
Answer: D O= 21.54 miles.
Tasks for calculating visibility ranges
A) the visible horizon ( D e) and landmark ( D P)
B) Lighthouse opening fire
conclusions
1. The main ones for the observer are:
a) planes:
The plane of the true horizon of the observer (pl. IGN);
The plane of the true meridian of the observer (pl. IMN);
The plane of the first vertical of the observer;
b) lines:
The plumb line (normal) of the observer,
Line of the true meridian of the observer ® noon line N-S;
Line E-W.
2. Direction counting systems are:
Circular (0°¸360°);
Semicircular (0°¸180°);
Quarter (0°¸90°).
3. Any direction on the surface of the Earth can be measured by an angle in the plane of the true horizon, taking the line of the true meridian of the observer as the origin.
4. True directions (IR, IP) are determined on the ship relative to the northern part of the true meridian of the observer, and KU (heading angle) - relative to the bow of the longitudinal axis of the ship.
5. Range of the visible horizon of the observer ( D e) is calculated by the formula:
.
6. The visibility range of a navigational landmark (daytime in good visibility) is calculated by the formula:
7. Range of visibility of the fire of a navigational landmark, according to its range ( D K) shown on the map is calculated by the formula:
, where .